Structure 1. Models of the particulate nature of matter

Classify matter by asking what particles are present, whether different elements are chemically bonded, and whether the composition is fixed. An element contains one type of atom and cannot be chemically broken down. A compound contains atoms of different elements bonded in a fixed ratio, so it has its own properties. A mixture contains two or more substances in no fixed ratio; the components are not chemically bonded to one another and retain properties that can be used for separation.

Choose a separation method in three linked moves: identify a physical-property difference, select the operation that exploits it, then state which component is recovered where. Use magnetism for a magnetic solid; filtration for an insoluble solid suspended in a fluid; crystallization or evaporation to recover a dissolved solid; and simple distillation to recover a solvent or separate liquids with a sufficiently wide boiling-point gap. Use fractional distillation when miscible liquids have closer boiling points. Chromatography separates through different relative attractions to the mobile and stationary phases. A multi-component mixture may require a sequence of methods.

Uniform appearance is not evidence of a compound: air and salt solution are homogeneous mixtures. Also, do not justify filtration by saying only that substances have different solubilities. The separated solid must be insoluble, and the filter works because its particles do not pass through the pores.

Diagnostic check: a particle box containing two unbonded particle types represents a mixture even if it looks uniform. For an unknown mixture, complete three columns—chosen method, physical-property reason, and observable recovered fraction. Reject any proposal that changes the substances chemically or cannot state where each component goes.

The kinetic molecular theory is a model that explains observable state properties using particle arrangement, movement, spacing, and attractions. Solid particles are closely packed and vibrate about fixed positions. Liquid particles remain close but change neighbours and flow past one another. Gas particles are far apart relative to their size and move freely, so a gas fills its container and is much more compressible.

Do not say that particles themselves melt, expand, or become a different substance. In a physical state change, the same particles adopt different motion and arrangement. For molecular substances, the energy change is associated with intermolecular attractions; it does not generally break the covalent bonds inside each molecule.

Temperature on the Kelvin scale is proportional to the average translational kinetic energy of particles. At the same Kelvin temperature, samples of different gases have the same average kinetic energy, although lighter particles have a higher typical speed than heavier particles. Use absolute temperature, not degrees Celsius, when comparing kinetic energies or temperature ratios.

Worked example — convert before interpreting

For $25.0\,^\circ\mathrm{C}$, substitute into the conversion: $T=25.0+273.15=298.15\,\mathrm{K}$, reported as $298.2\,\mathrm{K}$ to one decimal place. The Kelvin value is the absolute temperature used for kinetic-energy comparisons; a rise of $10\,\mathrm{K}$ is the same temperature interval as a rise of $10\,^\circ\mathrm{C}$.

Read a heating curve by first deciding whether the substance is within one state or changing state. On a sloping section, supplied energy increases average kinetic energy, so temperature rises. On a horizontal phase-change section at constant pressure, energy is still absorbed, but it is used to overcome intermolecular attractions and change particle arrangement; temperature and average kinetic energy stay constant until that change is complete. Cooling reverses the energy flow.

A 20 K temperature interval has the same size as a 20 °C interval, but 20 °C is not an absolute temperature of 20 K. Do not infer that heavier gas particles have greater average kinetic energy at the same temperature, and do not interpret a heating-curve plateau as a period when no energy is transferred.

An atom has a dense, positively charged nucleus containing protons and neutrons. Negatively charged electrons occupy the space outside the nucleus. Protons and neutrons are nucleons.

Read Z first to obtain protons, subtract Z from A to obtain neutrons, then use the ion charge to check or calculate electrons.

Apply the symbols in a fixed order. For ³⁵₁₇Cl⁻, Z = 17 gives 17 protons, A − Z gives 18 neutrons, and the 1− charge means one more electron than protons, so there are 18 electrons. The ion charge changes electron count, never the element identity.

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have the same electron arrangement and therefore the same chemical properties, while their different masses can give different physical properties.

Convert percentage abundances to fractions, multiply each isotope mass by its fractional abundance, then add the contributions. The result is a weighted mean, not an unweighted average.

For a two-isotope sample containing 75% mass 35 and 25% mass 37, the weighted mean is (35 × 0.75) + (37 × 0.25) = 35.5. An answer between the isotope masses is a useful check; simply averaging 35 and 37 would ignore abundance.

Read isotope identity from the pair of nuclear symbols: ³⁵₁₇Cl and ³⁷₁₇Cl are both chlorine because Z = 17 in each, but they contain 18 and 20 neutrons respectively because A differs. Mass number belongs to one nuclide; relative atomic mass is the abundance-weighted mean for a sample.

In a mass spectrum, peak position identifies an isotope's mass-to-charge value and relative peak height represents its abundance in the sample. Use those two pieces of evidence together to identify isotopes and determine relative atomic mass.

If singly charged isotope peaks occur at m/z 35 and 37 in a 3:1 height ratio, use fractional abundances 0.75 and 0.25 to obtain Aᵣ = 35.5. First confirm the charge state: m/z is a ratio, so a multiply charged ion cannot be read as mass alone.

An emission photon is released when an electron falls from a higher energy state to a lower energy state. Absorption moves an electron upward and requires photon energy.

Across electromagnetic radiation, shorter wavelength means higher frequency, and higher frequency means higher photon energy. Explain the electron direction and photon exchange when distinguishing absorption from emission.

Read every spectral transition in two directions: absorption raises an electron by ΔE, while a downward transition emits a photon with ΔE = hf = hc/λ. A shorter-wavelength line therefore represents a larger energy gap, not a higher line intensity.

Orient the spectrum before comparing lines: radio → microwave → infrared → visible → ultraviolet → X-ray → gamma is increasing frequency and photon energy, and decreasing wavelength. Within visible light, red has longer wavelength and lower photon energy than violet.

Hydrogen's emission spectrum contains discrete lines because electrons occupy discrete energy levels. Each line corresponds to a downward transition and the emitted photon's energy equals the energy difference between the levels.

Use the presence of separate lines as evidence for discrete levels, and use convergence at higher energy or frequency as evidence that the level spacing becomes smaller.

Compare lines by their energy gaps. Transitions ending at n = 2 form the visible series, and lines crowd together as the starting level rises because adjacent high-n levels are closer in energy. The convergence limit represents removal of the electron, not one more bound-state transition.

A main energy level, or shell, is identified by the principal quantum number n = 1, 2, 3, and so on.

For n = 3, the theoretical capacity is 2(3²) = 18 electrons. This is a capacity, not a claim that every third shell is full: the actual occupancy depends on the atom and the relative energies of available sublevels.

A main energy level contains only the sublevels allowed by its principal quantum number: n = 1 has s; n = 2 has s and p; n = 3 has s, p and d; and n = 4 can include s, p, d and f. Within one main level the sublevels rise in energy s < p < d < f, while the filling order across different levels can interleave, as the next card makes explicit. Each sublevel contains a fixed number of orbitals.

For recognition questions, keep the hierarchy clear: main energy level → sublevel → orbital. The s and p shapes are the explicitly required shape evidence here.

Use the hierarchy as a classification test: a p sublevel contains three orbitals, and each orbital can hold two electrons. An orbital describes a probability region with a characteristic shape; it is not a circular route travelled by an electron.

Aufbau fills lower-energy orbitals first. Pauli limits an orbital to two electrons with opposite spins. Hund's rule places electrons singly in degenerate orbitals before pairing.

Check total electrons, obey the filling order, apply Hund and Pauli in each sublevel, and treat the Cr/Cu exceptions explicitly rather than forcing the naive pattern.

Build an orbital diagram by checking electron total, energy order, single occupation of equal-energy orbitals, then opposite-spin pairing. For transition-metal ions remove 4s electrons before 3d, and verify Cr and Cu against the stated exceptions rather than forcing the simple filling pattern.

Worked ion check: Fe has 26 electrons and condensed configuration [Ar] 4s² 3d⁶. To form Fe³⁺, remove the two electrons from the highest principal level, 4s, before removing one 3d electron, giving [Ar] 3d⁵. The final superscripts total 23 electrons, matching 26 − 3.

First ionization energy is the energy required to remove one mole of electrons from one mole of gaseous atoms. The convergence limit in an emission spectrum corresponds to ionization.

Worked example — hydrogen convergence limit

The local course book gives $\lambda=9.12\times10^{-8}\,\mathrm{m}$. First, $f=c/\lambda=(3.00\times10^8\,\mathrm{m\,s^{-1}})/(9.12\times10^{-8}\,\mathrm{m})=3.29\times10^{15}\,\mathrm{s^{-1}}$. Then $E_{\text{photon}}=hf=(6.63\times10^{-34}\,\mathrm{J\,s})(3.29\times10^{15}\,\mathrm{s^{-1}})=2.18\times10^{-18}\,\mathrm{J}$. Convert one-photon energy to one mole and joules to kilojoules: $IE=(2.18\times10^{-18})(6.02\times10^{23})/1000=1.31\times10^3\,\mathrm{kJ\,mol^{-1}}$. This is the molar energy for the first ionization process $\mathrm{H(g)\rightarrow H^+(g)+e^-}$.

At the convergence limit, convert wavelength or frequency to energy per photon with E = hf, then multiply by the Avogadro constant and convert J mol⁻¹ to kJ mol⁻¹. Across-period trends are general patterns; subshell energy and electron pairing explain the named dips.

Successive ionization energies rise as electrons are removed. A very large increase occurs when removal crosses from the outer shell into a lower, more tightly held shell.

A jump after three outer-electron removals is the pattern used in the study guide for a group 13 element. The same count-before-the-jump method applies to new data.

Locate the first order-of-magnitude jump before naming the group. A large jump after the second electron is removed shows two outer-shell electrons; the third electron would come from a lower shell, supporting a group 2 assignment.

The mole is the SI unit for amount of substance. One mole contains the Avogadro constant number of specified entities, such as atoms, molecules, ions, or electrons.

Identify which entity the question asks for, then multiply the amount in moles by Avogadro's constant and by the number of those entities in each formula unit or molecule when needed.

Specify the entity before calculating. One mole of H₂O contains one mole of molecules, two moles of H atoms and one mole of O atoms; multiplying by Nₐ without the formula-unit multiplier answers a different particle question.

Relative atomic mass is a weighted mean relative to one twelfth of the mass of a carbon-12 atom. Relative formula mass is the sum of the relative atomic masses represented in the formula.

Relative atomic mass and relative formula mass are ratios on the carbon-12 scale, so they have no units. Apply every subscript, including waters of crystallization or repeated ions.

For Ca(OH)₂, include both bracketed groups: Mᵣ = Aᵣ(Ca) + 2[Aᵣ(O) + Aᵣ(H)]. Keep Mᵣ dimensionless; attach g mol⁻¹ only when the same numerical total is used as a molar mass.

Molar mass is the mass of one mole of a substance, measured in g mol⁻¹. It connects mass, amount, and particle number in a conversion chain.

To reach particles from mass, divide mass by molar mass to obtain moles, then use the Avogadro conversion and count the requested entities if the formula contains more than one.

Use units to choose the direction: 9.0 g of H₂O divided by 18.0 g mol⁻¹ gives 0.50 mol. To find molecules, continue from moles to nNₐ; do not multiply mass directly by the Avogadro constant.

An empirical formula gives the simplest whole-number ratio of atoms. A molecular formula gives the actual number of each atom in a molecule.

Keep the empirical ratio simplest, and make sure the molecular-formula multiplier is a whole number consistent with the given molar mass.

A composition of 40.0% C, 6.7% H and 53.3% O gives the simplest ratio CH₂O after division by atomic masses. If the molar mass is 180 g mol⁻¹, compare it with the empirical-formula mass 30 to obtain the multiplier 6 and molecular formula C₆H₁₂O₆.

Do not round a ratio such as 1 : 1.50 : 1 to 1 : 2 : 1. Preserve the calculated values and multiply every ratio by the same small integer: ×2 converts halves, while ×3 can resolve values close to thirds such as 1.33 or 1.67. Round only after the common multiplier produces values consistent with the data precision.

Molar concentration is the amount of solute in moles per cubic decimetre of solution. Square brackets can denote molar concentration.

Use V in dm³ when calculating C in mol dm⁻³. Rearrange the same relationship to find amount or volume, and keep the solution volume distinct from the solute mass.

Convert volume before substitution: 500 cm³ = 0.500 dm³, so 0.250 mol in that solution gives C = 0.500 mol dm⁻³. The denominator is the final solution volume, not the volume or mass of solute alone.

Adding solvent increases the final solution volume while the amount of solute stays constant, so concentration decreases. Use the final solution volume—not the solvent volume added—and do not apply c₁V₁ = c₂V₂ when solute reacts or is removed.

At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules. Under the same conditions, gas-volume ratios follow mole ratios.

Balance the equation, identify the limiting gas amount when needed, apply the coefficient ratio to gas volumes at the same temperature and pressure, then convert to the requested volume units.

For N₂ + 3H₂ → 2NH₃ at one temperature and pressure, one gas volume of N₂ requires three equal gas volumes of H₂ and forms two of NH₃. This direct volume ratio works because the gases share conditions; otherwise convert through moles.

An ideal gas is a simplified particle model. Use its assumptions as a checklist before deciding whether PV=nRT is a suitable description of a real sample.

Use the assumptions to make predictions: compressing a gas until particle volume is no longer negligible weakens the model, while raising temperature usually reduces the relative importance of attractions. Ideal particles still move and collide; only the collisions are treated as elastic.

Gas pressure arises from particle collisions with the container walls and the associated momentum transfer. At a higher Kelvin temperature, the greater average kinetic energy established in Structure 1.1.3 changes the collision behaviour; this does not replace the separate assumptions that particle volume and intermolecular attractions are negligible.

Real gases deviate most from ideal behaviour at low temperature and high pressure. Low temperature reduces particle kinetic energy, while high pressure brings particles close together.

A strong explanation names the condition, identifies the failed ideal assumption, and links it to the observed deviation.

Diagnose the cause from the condition. Cooling makes attractive forces more important because particle kinetic energy is lower; strong compression exposes both attractions and finite particle volume. Name the failed ideal assumption rather than stating only that the gas is 'non-ideal'.

Molar volume is the volume occupied by one mole of gas at a specified temperature and pressure. At STP as used by the current IB data context (273.15 K and 100 kPa), Vₘ is approximately 22.7 dm³ mol⁻¹; a different condition requires its own value.

Read pressure–volume graphs as an inverse relationship, not simply as one quantity increasing while the other decreases. At STP, use the stated molar volume with the balanced-equation mole ratio.

Attach conditions to every molar volume. Once Vₘ is valid for the stated temperature and pressure, convert gas volume to moles, apply the balanced-equation ratio, then convert back if needed. Do not carry one tabulated Vₘ into a different set of conditions.

Worked example — amount and molar mass from gas data

The local course book gives $P=101.3\,\mathrm{kPa}$, $V=1.91\,\mathrm{dm^3}$, $m=3.30\,\mathrm{g}$ and $T=150\,^\circ\mathrm{C}=423.15\,\mathrm{K}$. Because $\mathrm{kPa\,dm^3=J}$, use $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$: $n=PV/(RT)=(101.3\times1.91)/(8.31\times423.15)=0.0550\,\mathrm{mol}$. Then $M=m/n=3.30\,\mathrm{g}/0.0550\,\mathrm{mol}=60.0\,\mathrm{g\,mol^{-1}}$. The final value is the mass of one mole of the vaporized compound under the stated ideal-gas model.

Convert Celsius to kelvin before substitution, and make pressure and volume units consistent with the chosen gas constant. Rearrange the equation only after the known quantities and units are identified.

Before solving, write a unit line beside P, V and T. With R = 8.31 J mol⁻¹ K⁻¹, use pressure in Pa, volume in m³ and temperature in K; using kPa with dm³ is also consistent because kPa·dm³ equals J. Judge model suitability before trusting the numerical result.